 Consider the three points illustrated on the screen. Point A, which has the coordinates negative 2, positive 1, B, which has the coordinates 2, 3, and C, which has the coordinates 3, 1. These points are clearly non-colinear, and therefore they form a triangle. What we want to first do is verify that this triangle ABC is in fact a right triangle. How do we show that as a right angle? And then once it determines the right triangle, what we want to do is actually compute the area of the triangle. So how can we show that a triangle is right? Well, this is one of the goals of trigonometry, but without using trigonometric ratios like sine, cosine, and tangent, can we do this just from the coordinates in the formulas we know already? It turns out we can do this from the Pythagorean equation because looking at the picture here, it seems that this angle B is a right angle. So can we show that this is the hypotenuse of the triangle using the Pythagorean equation? That is, if this is point A, this is point B, this is point C, can we show that A squared plus C squared equals B squared? That's what we'd want to show. And so let's compute the distances of these lengths of the triangle. That is, find the triangle sides. So let's first find the distance AB. By the distance formula, we're going to get 2 minus a negative 2, so that's a 2 plus 2 squared. So we took the difference of the x-coordinate and squared it, and then we're going to add that to the distance of the difference of the y-coordinate, 3 minus 1 squared. So continuing on, you get 2 plus 2, which is 4, 4 squared is 16, 3 minus 1 is 2, 2 squared is a 4. So we get the square root of 20, right? For which, if you want, you can simplify this radical because even though 20 itself is not a perfect square, it does have a perfect square divisor. 20 factors as 4 times 5, which of course, 4 is a perfect square. And so this is the same thing as 2 root 5. We don't need an approximate value. We'll just be content with the square root of 20 or 2 root 5. So that's the length of AB, this distance right here. So we could write that as 2 root 5. What about the distance, say, AC? So again, using the distance formula, we're going to end up with doing the following. Take the difference of the x-coordinate, so we're going to get 3 minus negative 2, which is 3 plus 2 squared. Add to that the difference of the y-coordinate squared, so you get 1 minus 1 squared. So in this situation, you end up with a 5 squared plus 0, that is, this is just the square root of 5 squared. I don't need to know that 5 squared is 25 because when I take the square root, I'm going to get back a 5. So the distance between A and C is actually a whole number. It's going to be 5. And so let's do the last distance here. Let's do BC. Again, using the distance formula, you're going to get the square root here. Take the difference of the x-coordinate squared, so you're going to get 3 minus 2 squared plus, now take the difference of the y-coordinate 1 minus 3 squared. And it doesn't matter who's first or who's second because in the end, when you take the square of a negative, it'll be positive anyways. So 3 minus 2 is 1. 1 squared is 1. 1 minus 3 is negative 2 when you square that you're going to get a positive 4. And so we get that the distance between B and C is the square root of 5. So this distance right here is twice as long as this one right here. So now that we have these distances, now let's check the Pythagorean equation. So if we take 2 times the square root of 5 squared plus the square root of 5 squared, is that equal to the square root of 5 squared? Well, let's see that. Well, on the left-hand side, you're going to get 2 times the square root of 5. So that's going to be 4 times 5. Because 2 times 2 is 4. The square root of 5 times the square root of 5 is 5. And then over here, when you take the square root of 5, you square, you're going to get a 5. 5 squared we know is 25. Are these equal to each other? We're not sure yet. 4 times 5 is 20. In fact, we made that observation earlier, right? It's the square root of 20. So you get 20 plus 5. That actually does equal 25. We can verify that. And so since the three line segments do satisfy the Pythagorean equation, this is in fact a right triangle. So that's one way you could check to see if it triangles right or not. Another approach you could take was to make an argument using slopes. I want to present both types of arguments here for your benefit. So if this is in fact a right angle, that means the slope of this line and this line, their slopes must be negative reciprocals. That is the product of the slopes must be negative 1. And so we could actually identify what those are. Just do classic rise over run. So we go up, we go over. What is that? Well, as you go from A to B, you go up by two units. Notice take the difference of three and one is two. And then that's your rise. Then your run is you're going to go over four units. Notice I just took two minus negative two. So you get four right there. So rise over run, we end up with two fourths, which simplifies to be one half. Let's consider the slope of the line BC. So if we do our run and our rise, it doesn't matter which direction you draw them. Just, just, just pick one. We could also go down like this and over like that. But I'm going to do mine so it's exterior to the triangle diagram. So the run, that is the change of x coordinates. If you take the difference three minus two, you end up with a one. And then if you look at the rise, you're going to go from y equals three down to y equals one. So that's going to give you a goes down by negative two. So in that slope, you get rise over run, which is negative two. Notice that these slopes are negative reciprocals of each other, one half and negative two. If you take one half and you times it by negative two, you end up with negative one. That shows that the lines are perpendicular. So this shows you two different ways you could do to show that two different methods you could do to show that the triangle, excuse me, is right. Now that we know the triangle is right, we need to compute its area. So the area, the area of the triangle, this is going to be the classic one half base times height. The advantage of knowing that the triangle is a right triangle is that the base and the height are going to be the legs of the triangle. When your triangle is a non-right triangle, that is an oblique triangle, this becomes a little bit more difficult. And that's something we'll tackle a little bit later on in this lecture series. So this, the area is going to be one half times the distance AB times the distance BC, which we saw that AB turned out to be two root five. And we saw that BC turned out to be the square root of five. And so what happens is one half times two is equal to one, and the square root of five times the square root of five turns out to be five. So the area of this right triangle is here five. So I presented you two methods. You think the Pythagorean equation directly seemed a little bit more cumbersome because you have to go through all of these distance calculations. Then you have to check the formula. But just so you know, computing these slopes does the basic same calculation that the distance formula did. You have to calculate the ratio of the differences of x-coordinates and y-coordinates. You do that here in the distance formula. There's just a little bit extra that you would do. And to be aware that find the area, we have to find these distances anyways. So those calculations are not redundant. But again, two different approaches to the same problem.